BE Civil Engineering (IOE, TU) Engineering Mathematics III (IOE, SH 501) Question Paper 2076 Nepal

Setting up the region. The curves $y=x^2$ and $y=x$ intersect where $x^2=x \Rightarrow x(x-1)=0$, so $x=0$ and $x=1$. For $0\le x\le 1$, the line $y=x$ lies above the parabola $y=x^2$. The region is

$$R=\{(x,y): 0\le x\le 1,\; x^2\le y\le x\}.$$

Green's theorem. For $P=3x^2-8y^2$ and $Q=4y-6xy$,

$$\oint_C P\,dx + Q\,dy = \iint_R\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dA.$$

Compute the partials:

$$\frac{\partial Q}{\partial x}=-6y,\qquad \frac{\partial P}{\partial y}=-16y,\qquad \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=-6y+16y=10y.$$

Evaluating the double integral.

$$\iint_R 10y\,dA=\int_0^1\!\!\int_{x^2}^{x} 10y\,dy\,dx=\int_0^1 10\cdot\frac{y^2}{2}\Big|_{x^2}^{x}dx=\int_0^1 5\left(x^2-x^4\right)dx.$$ $$=5\left[\frac{x^3}{3}-\frac{x^5}{5}\right]_0^1=5\left(\frac13-\frac15\right)=5\cdot\frac{2}{15}=\frac{10}{15}=\frac23.$$

Direct verification (along the boundary). The boundary consists of $C_1$: $y=x^2$ from $(0,0)$ to $(1,1)$, then $C_2$: $y=x$ from $(1,1)$ back to $(0,0)$.

Along $C_1$ ($y=x^2,\ dy=2x\,dx,\ x:0\to1$):

$$P\,dx+Q\,dy=(3x^2-8x^4)\,dx+(4x^2-6x\cdot x^2)(2x)\,dx=(3x^2-8x^4)\,dx+(8x^3-12x^4)\,dx.$$ $$\int_0^1(3x^2-8x^4+8x^3-12x^4)\,dx=\int_0^1(3x^2+8x^3-20x^4)\,dx=\Big[x^3+2x^4-4x^5\Big]_0^1=1+2-4=-1.$$

Along $C_2$ ($y=x,\ dy=dx,\ x:1\to0$):

$$P\,dx+Q\,dy=(3x^2-8x^2)\,dx+(4x-6x^2)\,dx=(-5x^2+4x-6x^2)\,dx=(4x-11x^2)\,dx.$$ $$\int_1^0(4x-11x^2)\,dx=\Big[2x^2-\tfrac{11}{3}x^3\Big]_1^0=0-\left(2-\tfrac{11}{3}\right)=-\left(\tfrac{6-11}{3}\right)=\tfrac53.$$

Total (direct). $-1+\dfrac53=\dfrac{-3+5}{3}=\dfrac23.$

Both methods agree.

$$\boxed{\oint_C (3x^2-8y^2)\,dx+(4y-6xy)\,dy=\frac23}$$

Divergence theorem. If $V$ is a closed region in space bounded by a piecewise-smooth closed surface $S$ with outward unit normal $\hat n$, and $\vec F$ has continuous first partial derivatives on $V$, then

$$\iint_S \vec F\cdot\hat n\,dS=\iiint_V (\nabla\cdot\vec F)\,dV.$$

Compute the divergence. With $\vec F=(4xz,\,-y^2,\,yz)$,

$$\nabla\cdot\vec F=\frac{\partial}{\partial x}(4xz)+\frac{\partial}{\partial y}(-y^2)+\frac{\partial}{\partial z}(yz)=4z-2y+y=4z-y.$$

Integrate over the unit cube $0\le x,y,z\le 1$:

$$\iiint_V(4z-y)\,dV=\int_0^1\!\!\int_0^1\!\!\int_0^1(4z-y)\,dx\,dy\,dz.$$

The integrand is independent of $x$, so the $x$-integral contributes a factor $1$:

$$=\int_0^1\!\!\int_0^1(4z-y)\,dy\,dz.$$

Integrate over $y$:

$$\int_0^1(4z-y)\,dy=\left[4zy-\frac{y^2}{2}\right]_0^1=4z-\frac12.$$

Integrate over $z$:

$$\int_0^1\left(4z-\frac12\right)dz=\left[2z^2-\frac{z}{2}\right]_0^1=2-\frac12=\frac32.$$

Result.

$$\boxed{\iint_S\vec F\cdot\hat n\,dS=\frac32=1.5}$$

The positive value indicates a net outward flux of $1.5$ units across the cube's surface.

Take the Laplace transform. Let $Y(s)=\mathcal{L}\{y(t)\}$. Using

$$\mathcal{L}\{y''\}=s^2Y-sy(0)-y'(0),\qquad \mathcal{L}\{y'\}=sY-y(0),\qquad \mathcal{L}\{e^{3t}\}=\frac{1}{s-3},$$

the equation becomes

$$\big(s^2Y-2s-3\big)-3\big(sY-2\big)+2Y=\frac{4}{s-3}.$$

Group the $Y$ terms:

$$(s^2-3s+2)Y-2s-3+6=\frac{4}{s-3}\;\Rightarrow\;(s^2-3s+2)Y=\frac{4}{s-3}+2s-3.$$

Since $s^2-3s+2=(s-1)(s-2)$,

$$Y=\frac{4}{(s-3)(s-1)(s-2)}+\frac{2s-3}{(s-1)(s-2)}.$$

Partial fractions, term 1:

$$\frac{4}{(s-3)(s-1)(s-2)}=\frac{A}{s-3}+\frac{B}{s-1}+\frac{C}{s-2}.$$

• $s=3$: $A=\dfrac{4}{(2)(1)}=2.$
• $s=1$: $B=\dfrac{4}{(-2)(-1)}=2.$
• $s=2$: $C=\dfrac{4}{(-1)(1)}=-4.$

So term 1 $=\dfrac{2}{s-3}+\dfrac{2}{s-1}-\dfrac{4}{s-2}.$

Partial fractions, term 2:

$$\frac{2s-3}{(s-1)(s-2)}=\frac{D}{s-1}+\frac{E}{s-2}.$$

• $s=1$: $D=\dfrac{2-3}{1-2}=\dfrac{-1}{-1}=1.$
• $s=2$: $E=\dfrac{4-3}{2-1}=1.$

So term 2 $=\dfrac{1}{s-1}+\dfrac{1}{s-2}.$

Combine:

$$Y=\frac{2}{s-3}+\left(2+1\right)\frac{1}{s-1}+\left(-4+1\right)\frac{1}{s-2}=\frac{2}{s-3}+\frac{3}{s-1}-\frac{3}{s-2}.$$

Inverse transform (using $\mathcal{L}^{-1}\{1/(s-a)\}=e^{at}$):

$$\boxed{y(t)=2e^{3t}+3e^{t}-3e^{2t}}$$

Check at $t=0$: $y(0)=2+3-3=2$ ✓. And $y'(t)=6e^{3t}+3e^{t}-6e^{2t}$, so $y'(0)=6+3-6=3$ ✓.

Cauchy residue theorem. If $f(z)$ is analytic inside and on a simple closed contour $C$ except for a finite number of isolated singularities $z_1,\dots,z_n$ interior to $C$, then

$$\oint_C f(z)\,dz=2\pi i\sum_{k=1}^{n}\operatorname{Res}_{z=z_k}f(z).$$

Locate singularities. The denominator $z^2-2z=z(z-2)$ vanishes at $z=0$ and $z=2$. Both are simple poles, and both lie inside $|z|=3$ (since $|0|=0<3$ and $|2|=2<3$).

Residue at $z=0$ (simple pole):

$$\operatorname{Res}_{z=0}=\lim_{z\to0}z\cdot\frac{z^2+1}{z(z-2)}=\lim_{z\to0}\frac{z^2+1}{z-2}=\frac{0+1}{0-2}=-\frac12.$$

Residue at $z=2$ (simple pole):

$$\operatorname{Res}_{z=2}=\lim_{z\to2}(z-2)\cdot\frac{z^2+1}{z(z-2)}=\lim_{z\to2}\frac{z^2+1}{z}=\frac{4+1}{2}=\frac52.$$

Apply the theorem.

$$\oint_C f(z)\,dz=2\pi i\left(-\frac12+\frac52\right)=2\pi i\cdot 2=4\pi i.$$ $$\boxed{\oint_C \frac{z^2+1}{z^2-2z}\,dz=4\pi i}$$

Assume a separated solution. Let $u(x,t)=X(x)\,T(t)$. Substituting into $u_t=c^2u_{xx}$:

$$X\,T'=c^2X''\,T\;\Rightarrow\;\frac{T'}{c^2T}=\frac{X''}{X}=-\lambda,$$

where $-\lambda$ is the separation constant (chosen negative so the time part decays). This gives two ODEs:

$$X''+\lambda X=0,\qquad T'+c^2\lambda T=0.$$

Solve the spatial problem with the boundary conditions. The conditions $u(0,t)=0$ and $u(L,t)=0$ force $X(0)=0$ and $X(L)=0$.

• If $\lambda\le 0$, only the trivial solution satisfies both BCs. So take $\lambda>0$, write $\lambda=k^2$.
• General solution: $X(x)=A\cos kx+B\sin kx$.
• $X(0)=0\Rightarrow A=0$, leaving $X=B\sin kx$.
• $X(L)=0\Rightarrow B\sin kL=0$. Nontrivially, $\sin kL=0\Rightarrow kL=n\pi$, so

$$k_n=\frac{n\pi}{L},\qquad \lambda_n=\left(\frac{n\pi}{L}\right)^2,\qquad n=1,2,3,\dots$$

Eigenfunctions: $X_n(x)=\sin\dfrac{n\pi x}{L}$.

Solve the time equation for each $\lambda_n$:

$$T_n'+c^2\lambda_n T_n=0\;\Rightarrow\;T_n(t)=B_n\,e^{-c^2(n\pi/L)^2 t}.$$

Superpose. By linearity the general solution is

$$\boxed{u(x,t)=\sum_{n=1}^{\infty}B_n\,\sin\!\frac{n\pi x}{L}\;e^{-c^2(n\pi/L)^2 t}}$$

Apply the initial condition. Setting $t=0$:

$$u(x,0)=\sum_{n=1}^{\infty}B_n\sin\frac{n\pi x}{L}=f(x).$$

This is the Fourier sine series of $f$ on $(0,L)$, so the coefficients are

$$\boxed{B_n=\frac{2}{L}\int_0^L f(x)\,\sin\frac{n\pi x}{L}\,dx,\qquad n=1,2,3,\dots}$$

Each mode decays in time at rate $c^2(n\pi/L)^2$; higher harmonics ($n$ large) decay fastest, so the temperature profile smooths and tends to $0$ as $t\to\infty$.

Convert to polar coordinates. Let $x=r\cos\theta$, $y=r\sin\theta$, so $x^2+y^2=r^2$ and $dA=r\,dr\,d\theta$. The first-quadrant quarter-disk of radius $2$ is described by

$$0\le r\le 2,\qquad 0\le\theta\le\frac{\pi}{2}.$$

The integral becomes

$$\iint_R e^{-(x^2+y^2)}\,dA=\int_0^{\pi/2}\!\!\int_0^{2} e^{-r^2}\,r\,dr\,d\theta.$$

Radial integral. Substitute $u=r^2$, $du=2r\,dr$, so $r\,dr=\tfrac12\,du$; limits $r:0\to2$ give $u:0\to4$:

$$\int_0^{2}e^{-r^2}r\,dr=\frac12\int_0^{4}e^{-u}\,du=\frac12\Big[-e^{-u}\Big]_0^{4}=\frac12\left(1-e^{-4}\right).$$

Angular integral.

$$\int_0^{\pi/2}d\theta=\frac{\pi}{2}.$$

Combine.

$$\iint_R e^{-(x^2+y^2)}\,dA=\frac{\pi}{2}\cdot\frac12\left(1-e^{-4}\right)=\frac{\pi}{4}\left(1-e^{-4}\right).$$

Numerically, $e^{-4}\approx 0.018316$, so the value $\approx \dfrac{\pi}{4}(0.981684)\approx 0.7711.$

$$\boxed{\iint_R e^{-(x^2+y^2)}\,dA=\frac{\pi}{4}\left(1-e^{-4}\right)\approx 0.771}$$

(a) Gradient of $\phi$. With $\phi=x^2y+y^2z+z^2x$,

$$\nabla\phi=\left(\frac{\partial\phi}{\partial x},\frac{\partial\phi}{\partial y},\frac{\partial\phi}{\partial z}\right)=\big(2xy+z^2,\;x^2+2yz,\;y^2+2zx\big).$$

At $(1,2,-1)$:

$$\frac{\partial\phi}{\partial x}=2(1)(2)+(-1)^2=4+1=5,$$ $$\frac{\partial\phi}{\partial y}=1^2+2(2)(-1)=1-4=-3,$$ $$\frac{\partial\phi}{\partial z}=2^2+2(-1)(1)=4-2=2.$$ $$\boxed{\nabla\phi\big|_{(1,2,-1)}=5\,\hat i-3\,\hat j+2\,\hat k}$$

(b) Divergence of $\vec F=(xy,\,yz,\,zx)$.

$$\nabla\cdot\vec F=\frac{\partial}{\partial x}(xy)+\frac{\partial}{\partial y}(yz)+\frac{\partial}{\partial z}(zx)=y+z+x.$$ $$\boxed{\nabla\cdot\vec F=x+y+z}$$

Curl of $\vec F$.

$$\nabla\times\vec F=\begin{vmatrix}\hat i&\hat j&\hat k\\[2pt]\dfrac{\partial}{\partial x}&\dfrac{\partial}{\partial y}&\dfrac{\partial}{\partial z}\\[4pt]xy&yz&zx\end{vmatrix}.$$

• $\hat i$: $\dfrac{\partial(zx)}{\partial y}-\dfrac{\partial(yz)}{\partial z}=0-y=-y.$
• $\hat j$: $-\Big(\dfrac{\partial(zx)}{\partial x}-\dfrac{\partial(xy)}{\partial z}\Big)=-(z-0)=-z.$
• $\hat k$: $\dfrac{\partial(yz)}{\partial x}-\dfrac{\partial(xy)}{\partial y}=0-x=-x.$

$$\boxed{\nabla\times\vec F=-y\,\hat i-z\,\hat j-x\,\hat k}$$

(c) Conservative? A field is conservative on a simply connected domain iff $\nabla\times\vec F=\vec 0$ everywhere. Here $\nabla\times\vec F=(-y,-z,-x)$, which is not identically zero (e.g. at $(1,2,-1)$ it equals $(-2,1,-1)\ne\vec0$). Therefore $\vec F$ is NOT conservative.

(a) Transform of $t e^{-2t}\sin 3t$. Start from $\mathcal{L}\{\sin 3t\}=\dfrac{3}{s^2+9}.$

Apply the first shifting theorem $\mathcal{L}\{e^{-2t}g(t)\}=G(s+2)$:

$$\mathcal{L}\{e^{-2t}\sin 3t\}=\frac{3}{(s+2)^2+9}.$$

Apply the multiplication-by-$t$ rule $\mathcal{L}\{t\,h(t)\}=-\dfrac{d}{ds}H(s)$ with $H(s)=\dfrac{3}{(s+2)^2+9}$:

$$\mathcal{L}\{t e^{-2t}\sin 3t\}=-\frac{d}{ds}\left[\frac{3}{(s+2)^2+9}\right]=-3\cdot\frac{-2(s+2)}{\big[(s+2)^2+9\big]^2}=\frac{6(s+2)}{\big[(s+2)^2+9\big]^2}.$$ $$\boxed{\mathcal{L}\{t e^{-2t}\sin 3t\}=\frac{6(s+2)}{\big[(s+2)^2+9\big]^2}}$$

(b) Inverse transform. Complete the square in the denominator:

$$s^2+4s+13=(s+2)^2+9.$$

Rewrite the numerator about $(s+2)$:

$$s+5=(s+2)+3.$$

So

$$\frac{s+5}{(s+2)^2+9}=\frac{s+2}{(s+2)^2+3^2}+\frac{3}{(s+2)^2+3^2}.$$

Using $\mathcal{L}^{-1}\!\left\{\dfrac{s+a}{(s+a)^2+b^2}\right\}=e^{-at}\cos bt$ and $\mathcal{L}^{-1}\!\left\{\dfrac{b}{(s+a)^2+b^2}\right\}=e^{-at}\sin bt$ with $a=2,\ b=3$:

$$\boxed{\mathcal{L}^{-1}\left\{\frac{s+5}{s^2+4s+13}\right\}=e^{-2t}\cos 3t+e^{-2t}\sin 3t=e^{-2t}\big(\cos 3t+\sin 3t\big)}$$

Fourier cosine transform. With the convention

$$F_c(\omega)=\sqrt{\frac{2}{\pi}}\int_0^{\infty}f(x)\cos(\omega x)\,dx,$$

and $f(x)=1$ on $(0,a)$, $0$ otherwise,

$$F_c(\omega)=\sqrt{\frac{2}{\pi}}\int_0^{a}\cos(\omega x)\,dx=\sqrt{\frac{2}{\pi}}\left[\frac{\sin\omega x}{\omega}\right]_0^{a}=\sqrt{\frac{2}{\pi}}\,\frac{\sin a\omega}{\omega}.$$ $$\boxed{F_c(\omega)=\sqrt{\frac{2}{\pi}}\,\frac{\sin a\omega}{\omega}}$$

Inversion to deduce the integral. The inverse cosine transform recovers $f$:

$$f(x)=\sqrt{\frac{2}{\pi}}\int_0^{\infty}F_c(\omega)\cos(\omega x)\,d\omega=\frac{2}{\pi}\int_0^{\infty}\frac{\sin a\omega}{\omega}\cos(\omega x)\,d\omega.$$

Evaluate at $x=0$, where $f(0^+)=1$ (interior point of the unit step):

$$1=\frac{2}{\pi}\int_0^{\infty}\frac{\sin a\omega}{\omega}\,d\omega.$$

Hence

$$\int_0^{\infty}\frac{\sin a\omega}{\omega}\,d\omega=\frac{\pi}{2}\qquad(a>0).$$ $$\boxed{\int_0^{\infty}\frac{\sin a\omega}{\omega}\,d\omega=\frac{\pi}{2}}$$

This is the classical Dirichlet integral; its value is independent of $a$ as long as $a>0$ (for $a<0$ it equals $-\pi/2$, by oddness in $a$).

Harmonic check. Compute second partials of $u=x^3-3xy^2+2y$:

$$u_x=3x^2-3y^2,\quad u_{xx}=6x;\qquad u_y=-6xy+2,\quad u_{yy}=-6x.$$

Then $u_{xx}+u_{yy}=6x-6x=0$, so $u$ satisfies Laplace's equation and is harmonic.

Find the conjugate via Cauchy–Riemann. The CR equations are $u_x=v_y$ and $u_y=-v_x$.

From $v_y=u_x=3x^2-3y^2$, integrate with respect to $y$ (treat $x$ constant):

$$v=\int(3x^2-3y^2)\,dy=3x^2y-y^3+g(x),$$

where $g(x)$ is an unknown function of $x$.

Differentiate with respect to $x$: $v_x=6xy+g'(x)$. The second CR equation requires $v_x=-u_y=6xy-2$. Equate:

$$6xy+g'(x)=6xy-2\;\Rightarrow\;g'(x)=-2\;\Rightarrow\;g(x)=-2x+C.$$

Thus

$$\boxed{v(x,y)=3x^2y-y^3-2x+C.}$$

Express $f(z)$ in terms of $z$.

$$f=u+iv=(x^3-3xy^2)+i(3x^2y-y^3)+\big(2y-2ix\big)+iC.$$

Note $(x^3-3xy^2)+i(3x^2y-y^3)=(x+iy)^3=z^3$. Also $2y-2ix=-2i(x+iy)=-2iz$. Hence

$$\boxed{f(z)=z^3-2iz+iC.}$$

Check: $f'(z)=3z^2-2i$ exists everywhere, confirming $f$ is entire (analytic on all of $\mathbb{C}$).

(a) Classification. For $Au_{xx}+Bu_{xy}+Cu_{yy}=0$, examine the discriminant $B^2-4AC$. Here $A=1,\ B=4,\ C=3$:

$$B^2-4AC=16-4(1)(3)=16-12=4>0.$$

Since the discriminant is positive, the PDE is hyperbolic.

(b) d'Alembert's solution. The equation $u_{tt}=c^2u_{xx}$ has $c^2=4\Rightarrow c=2$. d'Alembert's formula for an infinite string is

$$u(x,t)=\frac{1}{2}\big[f(x+ct)+f(x-ct)\big]+\frac{1}{2c}\int_{x-ct}^{x+ct}g(\xi)\,d\xi,$$

where $f(x)=u(x,0)$ and $g(x)=u_t(x,0)$.

Here $f(x)=\sin x$ and $g(x)=0$, so the integral term vanishes:

$$u(x,t)=\frac12\big[\sin(x+2t)+\sin(x-2t)\big].$$

Using the sum-to-product identity $\sin P+\sin Q=2\sin\frac{P+Q}{2}\cos\frac{P-Q}{2}$ with $P=x+2t,\ Q=x-2t$:

$$u(x,t)=\frac12\cdot 2\sin x\cos 2t=\sin x\,\cos 2t.$$ $$\boxed{u(x,t)=\sin x\,\cos 2t}$$

Evaluate at $\left(\tfrac{\pi}{2},\tfrac{\pi}{4}\right)$.

$$u\!\left(\frac{\pi}{2},\frac{\pi}{4}\right)=\sin\frac{\pi}{2}\,\cos\!\left(2\cdot\frac{\pi}{4}\right)=\sin\frac{\pi}{2}\,\cos\frac{\pi}{2}=(1)(0)=0.$$ $$\boxed{u\!\left(\tfrac{\pi}{2},\tfrac{\pi}{4}\right)=0}$$

Check IC: at $t=0$, $u(x,0)=\sin x\cos 0=\sin x$ ✓; and $u_t=-2\sin x\sin 2t$, so $u_t(x,0)=0$ ✓.